Problem set thick lenses and contacts 2010
Formulas to remember
Equivalent power = F1+F2(t/n lens)F1F2
Use similar triangles and solve for the unknown variable for retinal image size problems
Remember to vertex powers over +/- 4D
Lacrimal lens acronym SAMFAP (steep add minus, flat add plus) in 1:1 ratio
Lateral magnification LM= nl’/n’l
1. A patient has an Rx of -15.00+3.00x090 and wants to be fit with rgp lenses.
Assume a vertex distance of 12mm. Her K readings are 44.00 /47.50 and we
plan on making the lens so that the base curve is 0.5D steeper than the
flattest K reading. What will be the contact lens power that you should
order?
Answer:
1) Take the spherical equivalent of the Rx = -13.50
2) Vertex this = -11.62
3) Compensate for the lacrimal lens (Steep add Minus/ Flat add Plus) -0.50
4) Final CL power will be -11.82
1. On a tangent screen at a distance of 1m from the eye, a circular scotoma 20 cm
in diameter is mapped out. What is the size of the scotoma on the retina? What is
the retinal image size if the scotoma is 10 degrees?
Answer: set up similar triangles, assume f’ is 17mm (schematic eye)
Image=(20)(17)/(100x h’)=3.4mm
Note assume that the schematic eye has a cornea to retina distance of 24mm and the
nodal point is a 17mm.
Note that the size of the scotoma decreases to 1.7mm for a scotoma of 10 degrees so
there is a linear relationship!
4. The angular width of the blind spot is 6 degrees (tan 6 degrees = 0.105). What is
the width of the blind spot on the retina?
A. Recall height image/height object = distance image/distance object. So,
x/.105=17/1=1.8mm
5. The size of a road sign is about 300cm. When the sign is 50 meteres away, what
is the size of the sign on the retina? Set up the same way: x/3.0=.17/50 so, Retinal
image size 1.02mm
6. What if the road sign is 100 meters away? It will be .51mm Again, note the linear
relationship.
7. You have a patient with aphakia and mild keratoconus who wants to be fit with
a soft lens. The K values are 46.00/50.00 . Your base curve choices are: 8.8, 8.4
and 9.0. Which would be the most appropriate base curve?
Answer: It would be 8.4. Remember that when you are talking about radians (standard
for base curves) , the higher the number the flatter the base curve. When you convert
radians to diopters (F=n’-n/r) the opposite is true; the higher the number the steeper the
cornea. So, a steeper cornea will need a base curve (in radians) with a lower number.
7. You have a lens system with the following parameters: n (lens) = 1.45,
suspended in water(n=1.33), lens thickness is 3.6mm, r1=6.0mm, r2=10.0mm. Find
the front and back surface powers of the lens, find the equivalent (true) power,
find the position of the principal planes.
Answer: F1=n’-n/r= 1.45-1.33/.006=+20.0D
F2=(1.33-1.45)/.010= +12.00 D
Equivalent power = F1+F2(t/n lens)F1F2= 20+12(.0036/1.45)(20)(12)= +31.40
Position of H= n water (t/n lens)(F2/Fequiv) = 1.33(.006/1.45)(12/31.40)=+1.26mm
Position of H’= n water (t/n lens)(F1/Fequiv) = 1.33 (.006/1.45)(20/31.40)=+2.10mm
8. You have a lens system with the following parameters: first lens power is
+15.00, second lens power is -3.00. Both lenses are in air separated by a distance
of 17cm. there is an object located one meter away from the front surface. Where
is the final image located?
A. L=n/l= -1D
L’=F+L=15-1=14, so l’=1/14 or +7.14 cm from the first lens
Now do the same thing for the second lens, using the real image as an object for the
second refracting surface. Remember to subtract 7.14 from the 17mm to get the new
object distance so you can calculate the incoming vergence.
17-7.14=9.86 is the value for l2
L2=n/l2=1.-9.86=-10.14D and L2’=F+L2=-10.14-3.00=-13.14 D so image position is 1/13.14 or -7.61cm (virtual image to the left of the second refracting element)
9. Your low vision patient reports that he can take the magnifier you gave him
last year (5x) and combine it with a WalMart magnifier that says 10X and he
can read his medicine bottles. However, he would like a single magnifier to
do the same thing. What type of magnifier do you think you should order
for him? Answer: 50X. Just obtain the product of the magnifiers together.
10. What is the total lateral magnification if you have the following system? A
glass rod of 1.5 index suspended in air with an object 50 centimeters in
front of the rod? The image position of the first object is at 18.75cm, the
object position for the second refracting surface is 3.75 cm and the final
image position is 2.22 cm
Answer: You need to remember that LM= nl’/n’l
LM first refracting surface is 1.0(18.75)/1.5(-50) = -0.25
LM second refracting surface is 1.50(2.22)/1.0/(3.75) = +0.89
LM total = (-0.25)(0.89) or -0.22 and the image is inverted (negative sign) and
minimized (less than one).
2. A lens has a front surface of +6.5 D, thickness of 4mm and back surface of -12D.
Index is 1.50. Calculate back vertex power and the equivalent (true) powers. What
happens if you switch the front and back powers?
A. Just plug numbers into the back vertex power formula and the equivalent power
formulas which are found in the powerpoint presentation.
Back vertex power is -5.39, equivalent power is -5.29 D
1) Effective power is used to indicate the change in lens power required when a
lens is moved from one position in front of the eye to another . The formula for
this is: Fv=Fa/1-dFa
A 12D myope refracted at vertex 12mm wants contacts. What is the required
power of the contact?
Answer: Fv=-12.00/1-.012(-12.0) = -10.52D
2) You have a patient with an Rx of -4.00 = - 1.50 x 180 that you fit with a spherical
RGP lens. It fits well, but the patient does not seem to have good acuity. Your overrefraction is =0.50 = -1.00 x 090. What is the problem?
The patient is experiencing lens flexure. Fix this by increasing the center thickness
of the lens or going with a less oxygen permeable material, which will be stiffer. One
hint is that the axis is 90 degrees away from the refractive axis.